q25 - funny q (1 Viewer)

[fizzwizz]

I think q25 is incorrect. Students were asked to draw the line of best fit thgrough the steepest part of the graph (i have no idea why but anyway). This gives an intercept value on x-axis of about 1.60 eV. In part b of q, studnets have to use this line to determine work function. Of course most students will use the 1.60 value obtained from line of best fit. And i'm pretty sure this is what the inept question setters would want anyway, given the spoecific tie-in with the line of best fit.

However this is completely wrong. If 1.60 eV is work func, then how is that current has already been produced at lower photon energies? Specifically, at 1.2 eV there is a current of 0.5 microamps - this is in the given data. So 1.6 eV cant be work function if current is already produced at somewhere between - say half-way - 0.9 and 1.4 eV.

Or am I wrong?

 

[JohnBuchner]

I think by "steepest part of the graph" they meant all of the graph that was not flat on the bottom, ie. from 1.2 eV upwards. I thought it was pretty confusing myself.

 

[DigitalFortress]

shit...I found the y intercept...shit...
I didn't think -10.8 photocurrent was possible

 

[fizzwizz]

you could well be right John - in fact it it is the only way it could be answered correctly to determine work func. But the question does say..." ....in the region where the photocurrentvaries greatest with photon energy" And its that biot which worries me. To decide when to draw a line or a curve of best fit is not that hard - if it looks like a straight line then it probablys is. To correctly answer the question, as you have said rightly, then all that is needed is a simple interpolation between data points 2 and 3.

I guess its just another example of deliberate confusion setting - not a test of physics.

 

[helper]

Firstly joining two points is not going to give you a line of best fit.
The 1.2eV reading could have been delibrately there as a piece of erroneus data, which is why lines of best fit are drawn.

 

[fizzwizz]

Hi Helper. You could be right about an erroneous data value.....but I doubt it. There could certainly be errors in reading a current value to a certain degree of accuracy, but certainly not an error in actually recording that a current actually existed if it didn't.......ie if current at 1.2 eV was zero, then why would a current of 0.5 mcroaamps be recorded? I was referring to data points 4, 5 and 6...three points not two. I am guessing that this is the section referred to in question...otherwise a line of best fit is definitely not appropriate for any other set of data popints, regardless of experimental inaccuracies.

 

[helper]

The way these experiments work is to use colour filters. Most of these filters have a peak wavelength but are not totally monochromatic. This means that you have stray currents form.

Experimental results for this experiment commonly show this curve.

http://www.physics.rutgers.edu/ugrad/labs/photoelectric_html_25230628.gif
http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/tourexp/photef1.gif
http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/tourexp/photef2.gif
http://www.oufusion.org.uk/images/PhotoElec1.gif
http://www.oufusion.org.uk/images/PhotoElec2.gif

The document below goes through other reasons it is curved.
http://www.physics.uq.edu.au/people/mcintyre/php/laboratories/download_file.php?eid=44

The slight objection I have to the question, which this is the second time they have brought this experiment into the HSC, is that they are almost saying the experiment is compulsory.

 

[redd]

hey helper, since haboozin posted up the paper you think you could draw the graphs for us?

 

[fizzwizz]

helper - thanks for that - Filters "have" a range of wavelengths - thats fine and I can accept that. I guess its just a case of moving the goalposts. The links you gave - thanks for those - were all of pd vrs current, not quite the same as in exam q but there you go

 

[helper]

Fizzwiz, the reason I posted voltage is that the normal method of plotting these curves, with calculations then to find the frequency but it is the same physics.

Redd, the Line of best fit, I would expect is as fizz described as the three middle points are a straight line.

 

[richz]

kool, helper, i certainly agree with ur answer, i got ~1.6eV

 

[haboozin]

kool, helper, i certainly agree with ur answer, i got ~1.6eV

i dont think there would be any correct answer for that question.
it said "from the line drawn on your graph" so whatever answer comes from the line u drew is right?

i didnt draw the line like that interpreted it different as the steepest part involving most of those that had a gradient of more than 0.

btw, for question C where it asks: "the experiment is repeated but with the intensity doubled" predict the result of this second experiment by drawing a second line on the graph"

well i think that u should start from the same WORK FUNCTION (since intensity doesn't change that sine E=hf) but the gradient should be doubled for the next line ?? is that right... i hope so

 

[fizzwizz]

yes - haboozin - youre right on both accounts. A curve parallel to given curve should also be OK but i'd be interested to see how that would be marked in terms of the specification that they wanted a line drawn.

The original reason why i posted this was to put up for discussion something i found odd. Some students might, given either the ideas that they have been taught or simply an understanding that work function relaes to onset of photocurrent pproduction, have found their solution a bit disconcerting. This is all especially the mor annoying since it was a mandatory prac in 2001, wasnt asked about at all, and then removed in 2002 review. I suspect because of financial retraints on schools - not in terms of difficulty since they have asked about it several times. The answer was never really in any doubt - the questions forced the issue so that wasnt really the point of my misgvings about the question. A bigger issue is how much of the photoelectric effect needs to be know? it isnt even obvious that eionsteins pe eqn should be known: the deliberate lack of clarity in the syllabus on this topic gives examiners free reign in what to ask.

 

[helper]

I doubt if they will accept a curve as the skills section asks you to identify when the use of a curve of best fit is appropriate. They also clearly asked for a line not a curve.

The sentiments about the question I can agree with you over but it wasn't even compulsory under the original 99 copy of the syllabus.
Sadly the exam is again deciding the syllabus, which is what wasn't suppose to happen.

 

[fizzwizz]

agree absolutely helper But I think it was a mandatory experiment - in original syllabus? Check doc with tracked changes from BOS. Anyway - thanks for your input - alwaysw good to see from another perspective.

 

[helper]

I did check already. Under the old syllabus
perform a first-hand investigation to demonstrate the photoelectric effect

That doesn't say it had to be that experiment. In the context of the syllabus a repeating of Hertz's experiment does that. It definitely did not require a quantitative analysis.

 

[fizzwizz]

Touche. But...touchy touchy.

 

[helper]

Wasn't trying to bite your head off, I just am abrupt with answers sometimes. This is more to do with the examiners trying to define the syllabus, rather than the syllabus defining the exam as it should be.